Traverse bearings — ABCD is a parallelogram with ∠BAD = 60°. If the bearing of AB is 30°, what is the whole-circle bearing of side CD?

Introduction / Context:
Parallelogram geometry implies that opposite sides are parallel and equal. In a traverse, bearings of parallel lines are either identical or differ by 180°, depending on direction of travel along the side. This problem asks for the bearing of CD given the bearing of AB in a regular parallelogram ABCD with a known interior angle at A.

Given Data / Assumptions:

• Parallelogram ABCD in order A→B→C→D.
• Bearing of AB = 30° (whole-circle).
• Opposite sides AB ∥ CD and AD ∥ BC.

Concept / Approach:

Because AB and CD are parallel, their direction lines are the same. However, when traversing from C to D, CD is oriented opposite to AB from A to B. Therefore, the bearing of CD is the back bearing of AB, i.e., AB + 180° (mod 360°). The given interior angle ∠BAD = 60° confirms a plausible acute angle between AB and AD but does not alter the parallelism conclusion for AB and CD.

Step-by-Step Solution:

Verification / Alternative check:

Sketch a simple rectangle (special case of a parallelogram) to visualize that the top side bearing is 180° opposite to the bottom side when traversed in the polygonal order.

Why Other Options Are Wrong:

90°/270° — correspond to east–west bearings unrelated to the given 30° base.
120° — would be a direction parallel to AD/BC rather than AB/CD.

Common Pitfalls:

Confusing forward and back bearings; assuming opposite sides always have identical numeric bearings without accounting for direction of travel.

Final Answer:

210°